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Reactive Systems: Modelling, Specification and Verification - Cs.ioc.ee

Reactive Systems: Modelling, Specification and Verification - Cs.ioc.ee

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42 CHAPTER 3. BEHAVIOURAL EQUIVALENCES<br />

different deadlock behaviour when made to interact with other parallel processes—<br />

a highly undesirable state of affairs.<br />

In light of the above example, we are forced to reject the law<br />

α.(P + Q) = α.P + α.Q ,<br />

which is familiar from the st<strong>and</strong>ard theory of regular languages, for our desired<br />

notion of behavioural equivalence. (Can you s<strong>ee</strong> why?) Therefore we n<strong>ee</strong>d to<br />

refine our notion of equivalence in order to differentiate processes that, like the two<br />

vending machines above, exhibit different reactive behaviour while still having the<br />

same traces.<br />

Exercise 3.2 (Recommended) A completed trace of a process P is a sequence<br />

α1 · · · αk ∈ Act ∗ (k ≥ 0) such that there exists a sequence of transitions<br />

P = P0<br />

α1 α2<br />

αk<br />

→ P1 → P2 · · · Pk−1 → Pk ,<br />

for some P1, . . . , Pk. The completed traces of a process may be s<strong>ee</strong>n as capturing<br />

its deadlock behaviour, as they are precisely the sequences of actions that may lead<br />

the process into a state from which no further action is possible.<br />

1. Do the processes<br />

<strong>and</strong><br />

(CA | CTM) \ {coin, coff<strong>ee</strong>, tea}<br />

(CA | CTM ′ ) \ {coin, coff<strong>ee</strong>, tea}<br />

defined above have the same completed traces?<br />

2. Is it true that if P <strong>and</strong> Q are two CCS processes affording the same completed<br />

traces <strong>and</strong> L is a set of labels, then P \ L <strong>and</strong> Q \ L also have the<br />

same completed traces?<br />

You should, of course, argue for your answers. <br />

3.3 Strong bisimilarity<br />

Our aim in this section will be to present one of the key notions in the theory of<br />

processes, namely strong bisimulation. In order to motivate this notion intuitively,<br />

let us reconsider once more the two processes CTM <strong>and</strong> CTM ′ that we used above<br />

to argue that trace equivalence is not a suitable notion of behavioural equivalence<br />

for reactive systems. The problem was that, as fully formalized in Exercise 3.2, the

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